Question

What's the opposite of log? I mean, if I was solving something like this: log y = 2, how would I find y?

Answer 1

@mathmate

Answer 2

The inverse of \(log_{10}(x)\) is \(10^x\).
The inverse of \(log_{e}(x)\) is \(e^x\).
For example,
y=\(log_{10}(x)\), then \(x=e^y\).
log(y)=4 then y=e^4 (for log base e)
log10(y)=1.5, then y=10^(1.5)
etc.

Answer 3

If my problem just says log, do I assume it is base 10?

Answer 4

*\(y=log_{e}(x)\), then \(x=e^y.\)

Answer 5

It depends on the level of your class. In universities, usually it's log_e. In high school, it's mostly log10.
I suggest you check what the teacher does in other situations.

Answer 6

hmm....my lessons say this:

log y = – 0.1523 + 0.0414x

Now to solve the equation for y, so we don’t have to deal with the logarithm. Start by exponentiating, then using properties of logarithms to solve for y-hat.

This could also be done using ln or natural logarithms. The only difference would be base e instead of base 10.

Answer 7

Another way to check, if you have done Calculus in your other courses, it's a good indication that log means log_e.

Answer 8

For regression lines, you can get the correct answer using log10 or log e, it's just the resulting equation would be different. I suggest you use ln since you're doing stats and other calculus stuff.

Answer 9

So it doesn't matter what I use?

Answer 10

And how would I use ln?

Answer 11

I remember learning this, but totally forget...

Answer 12

@mathmate

Answer 13

If you're doing regression, the difference between the two logs is just a constant multiplier, that's why the resulting equation will be different...oooh, the residues will be different too.
However, if you use each equation to get y_hat, they should be the same.
Try it for yourself!

Answer 14

Okay, that makes sense. But how do I use ln again?

Answer 15

is that the e thing you showed up above?

Answer 16

@mathmate

Answer 17

Depending on the data, you could be transforming data where x'=log(x), assuming the transformation linearizes the function.
The linear regression line would then be y=ax'+b.
To find x for a given value of y, you start from
y=ax'+b,
y=a log(x)+b
log(x)=(y-b)/a
x=e^((y-b)/a).
That's where the inverse might come in.

Answer 18

Okay, thanks. I'll try that.

Answer 19

You're welcome! :)